52

Considering many elementary functions have an antiderivative which is not elementary, why does this type of thing not also happen in differential calculus?

Rodrigo de Azevedo
  • 1
Ian Krasnow
  • 643
  • 6
    Looks more of a coincidence to me. The most significant thing is that we have a load of rules for derivatives, including derivative of composition, so this boild down to some classical functions like $\exp,$\ln,\sin,\cos$, etc. Those were often obtained from some practical (mostly physical) applications, and defined through some differential equations, which give you their derivatives. – lisyarus Mar 20 '17 at 09:03
  • 8
    @lisyarus The origin in physical phenomena and the definitions from differential equations are not arguments to justify the existence of closed-forms. –  Mar 20 '17 at 10:45
  • 2
    If we can differentiate $f(x)$ and $g(x)$ then we can usually easily differentiate $f(x)+g(x)$, $f(x)g(x)$, $\frac{f(x)}{g(x)}$ and $f(g(x))$. – Henry Mar 20 '17 at 17:19
  • Maybe because limits are easier to compute than infinite sums... – ThisIsNotAnId Mar 20 '17 at 19:37
  • 1
    I wonder if there's a deep reason (e.g., category theory related, perhaps) – MathematicsStudent1122 Mar 20 '17 at 19:55
  • @MathematicsStudent1122 I doubt any explanation can completely avoid some kind of discussion of what it means to apply the derivative or integral operators. – ThisIsNotAnId Mar 20 '17 at 20:23
  • I think this question might need a different answer. Most answers here just eventually end up proving that all elementary functions must have elementary derivatives. But that is because of the question, not the answers. I think it would help, if this description was not your intention, to illustrate what you mean by "why". Is simply proving what most of the answers below have done enough of an answer (which is perfectly good), or did you have something else in mind? – ThisIsNotAnId Mar 20 '17 at 20:39
  • 1
    @lisyarus Antidifferentiation also has practical physical applications. – user253751 Mar 20 '17 at 20:52
  • @MathematicsStudent1122: Yes, there is a deep reason, and it's given in my answer. – Mike Jones Sep 03 '17 at 15:41

6 Answers6

39

Just think of how we find those elementary functions:

  • We start with the constant functions, which have derivative $0$, and the identity function $f(x)=x$ which has derivative $1$.

  • We combine functions by means of addition, subtraction, multiplication, division, composition. For all of those cases we have explicit rules for the derivative.

  • We define new functions as the integral of other functions (e.g. $\ln x$ as integral of $1/x$). Obviously when deriving those we get back the function we started with.

  • We define functions as the inverse of another function. Again, we've got an explicit formula for derivatives of inverse functions.

Any function that cannot be defined by a chain of such operations (and also some which can, using the integration rule) we don't consider elementary.

So basically the reason is in the way we construct elementary functions. In some sense, one could say it is because of what functions we consider elementary.

Indeed, this hold not only for elementary functions; even most non-elementary functions we use are defined through such operations (in particular by integrals).

celtschk
  • 43,384
  • 5
    The question could be recast as "why are there differentiation rules" vs. "why are there no antidifferentiation rules". –  Mar 20 '17 at 10:41
  • 1
    Actually the question should not mention antiderivatives at all, since it threw off most other answers, as far as I can tell. I think the OP only mentioned antiderivatives as a contrast, not as the "meat" of the topic. In this sense, this answer seems hitting the mark very well. – AnoE Mar 20 '17 at 12:55
  • 2
    Obviously ln(x) is also an elementary function, because it is the inverse of the exponential function. I thought that any function which can be obtained without integration, and without resorting to power series can be considered elementary. – Ian Krasnow Mar 20 '17 at 22:40
  • @AnoE Yes, I understand Liouville's Theorem, I only mentioned antiderivatives for contrast. – Ian Krasnow Mar 20 '17 at 22:42
  • 3
    Weren't logarithms known before their relationship to $\frac{1}{x}$? – jpmc26 Mar 21 '17 at 01:59
  • @YvesDaoust Your reformulation can be answered by the fact that there are much more locally integrable functions compared to the differentiable ones. – Sebastian Bechtel Mar 21 '17 at 14:46
  • @SebastianBechtel: does this hold for functions with a closed-expression ? –  Mar 21 '17 at 15:35
  • Does what hold for functions with "closed-expression" (whatever this should mean exactly;)) – Sebastian Bechtel Mar 21 '17 at 15:50
  • @SebastianBechtel: what you say: "there are much more locally integrable functions compared to the differentiable ones". –  Mar 21 '17 at 15:58
  • @celtschk: excellent answer indeed. Now can we go deeper and try to understand why formal differentiation is possible in those cases ? I mean how is it possible that the derivative of a polynomial can be expressed as another polynomial (I am not asking a proof of the rule, which I know by heart, but a thought about closed-forms; for non-elementary functions, formal differentiation doesn't work in general). A crude answer could be "as some functions admit a closed-form derivative "by chance", we chose a set of functions that is closed under differentiation". –  Mar 21 '17 at 16:11
  • @YvesDaoust As it is not clear what "closed-expression" means, I cannot answer this additional question properly, but for example continuous functions or bounded functions are locally integrable and this already shows that this class is massively larger than the differentiable ones. – Sebastian Bechtel Mar 21 '17 at 18:11
  • @SebastianBechtel: Wikipedia will tell you. –  Mar 21 '17 at 18:30
25

The short answer is that we have differentiation rules for all the elementary functions, and we have differentiation rules for every way we can combine elementary functions (addition, multiplication, composition), where the derivative of a combination of two functions may be expressed using the functions, their derivatives and the different forms of combination.

Integration, on the other hand, neither has a direct rule for multiplication of two functions nor for composition of two functions. We can integrate the corresponding rules for differentiation and get something that looks like it (integration by parts and substitution), but it only works if you're lucky with what elementary functions are combined in what way.

You might say that there is a hope that there are rules out there, that we just haven't found them yet. This is not true; it's been proven that there are always integrals of elementary functions that are not elementary themselves (under most reasonable definitions of "elementary functions"). It's a deep result known as Liouville's theorem.

Arthur
  • 199,419
  • 3
    But why do we have rules in one case and not the other ? (Though in a way, Liouville's theorem says that the pattern of the differentiation rules prevents the existence of integration rules.) –  Mar 20 '17 at 10:49
  • @YvesDaoust That I do not know. – Arthur Mar 20 '17 at 10:56
  • This is what makes the question quite interesting. –  Mar 20 '17 at 11:20
  • It's why I called it "The short answer". – Arthur Mar 20 '17 at 11:23
  • See also this question. – Yvan Velenik Mar 20 '17 at 18:13
  • "The short answer is that we have differentiation rules for all the elementary functions" - Right, but the question is why is that that true? You literally just restated what was asked as a sentence. – BlueRaja - Danny Pflughoeft Mar 20 '17 at 21:02
  • @Arthur Intuitively this seems like it should cover all cases, but somehow this answer just feels unsatisfying, like I really want to see a deeper reason if that exists. So what you're saying is that differentiation can't create a non-elementary function from an elementary one because there is a small number of ways to combine these functions, and there is a rule for each one. – Ian Krasnow Mar 20 '17 at 22:34
  • @IanKrasnow I admit that this is not a full answer. I didn't submit it thinking that I had imparted any great wisdom, but rather a quick, intuitive account of the most immediate facts. Apparently it was popular. The whims of the SE community can be fickle some times. – Arthur Mar 20 '17 at 23:03
  • @YvesDaoust Basically the reason is that integration is fundamentally different from differentiation. Differentiation is local (requires the knowledge of function in an arbitrarily small neighbourhood of a point) while integration is global. Differentiation is algebraic (even though it is usually defined as a limit, it's still very simple in many senses) while integration is not, it is defined as some abstract inverse function or even as some huge limit over partial sums. ..cont.. – Anton Fetisov Mar 21 '17 at 19:28
  • ... Differentiation is defined in the same way in any dimension, while higher dimensional antiderivatives simply do not exist. There are both analytical and topological condition on the function too have "antiderivative" in higher dimensions. Basically the classical 1-dimensional correspondence between differentiation and integration is a stroke of luck coming from the simple structure of the real line: it can be essentially uniquely broken down into pieces and composed from them. – Anton Fetisov Mar 21 '17 at 19:33
  • @AntonFetisov: IMO, all these considerations hold for general functions, but I don't see the connection with closed-forms. In what way would locality influence the expressibility with a formula ? Formulas are global. –  Mar 21 '17 at 19:36
  • @YvesDaoust Locality matters because locally any (analytic, but we don't consider others here) function can be represented by an infinite power series. It is a homomorphism, mapping sums to sums, products to products and even composition to composition. It reduces the theory of functions basically to algebra and combinatorics, and it is obvious by induction that there in principle exist formulas, expressing derivatives of products and compositions via functions and their derivatives. All that fails for integrals. It's a bit circular, but explains the point. – Anton Fetisov Mar 21 '17 at 19:46
  • @AntonFetisov: sorry, I don't understand. Consider the function Gamma, analytical except at negative integers. Its derivative cannot be expressed in terms of the elementary functions and itself. –  Mar 21 '17 at 19:56
  • @YvesDaoust Gamma isn't defined as a composition of elemetary functions either. You were asking specifically why we should expect rules for differentiation but not integration of compositions. – Anton Fetisov Mar 21 '17 at 20:32
9

We can prove this:

From Wikipedia, an elementary function is one of the types:

  1. $\exp,\log,\operatorname{id}$
  2. Constant
  3. The sum of two elementary functions.
  4. The product of two elementary functions.
  5. The difference of two elementary functions.
  6. The quotient of two elementary functions.
  7. The composition of elementary functions.

We shall show that every case has an elementary derivative.

  1. $\mathrm D\exp = \exp$, which is of category (1). $\mathrm D\log x = x^{-1}$, which are elementary, with $x^{-1}$ being of category (6) -> (2) -> (1)
  2. The derivative of a constant is $0$, which is elementary of category (2).
  3. Follows from the linearity of differentiation: $\mathrm D(f+g) = \mathrm Df + \mathrm Dg$. Under the assumption that $f,g$ has elementary derivatives, the derivative of $f+g$ is elementary.
  4. Similar to (3), via the product rule.
  5. (2),(3),(4)
  6. Similar to (3), via the quotient rule.
  7. Chain rule.

Since every elementary function must belong to one of the categories above, it must have an elementary derivative.

Henricus V.
  • 18,694
  • Why does this work for derivatives and not for integrals ? –  Mar 21 '17 at 15:37
  • 2
    @YvesDaoust There are no rules of anti-differentiation. In particular, (4), (6), (7) fails to have elementary anti-derivatives in most cases. – Henricus V. Mar 21 '17 at 15:37
  • Of course, I know that (anyway we have the Risch algorithm). The question is why there is such an asymmetry. –  Mar 21 '17 at 15:49
  • @YvesDaoust For that you need to ask Liouville – Henricus V. Mar 21 '17 at 15:50
  • One can also wonder why formal differentiation works at all. For more exotic functions, such as Gamma, the derivative is out of reach. –  Mar 21 '17 at 15:51
  • @YvesDaoust The derivative of $\Gamma$ is yet another integral on $\mathbb R_{>0}$ and resembles the $\Gamma$ integral. – Henricus V. Mar 22 '17 at 13:49
  • Absolutely. And it is not expressible in terms of elementary functions nor Gamma. (By contrast with $x^n$ or $e^x$.) –  Mar 22 '17 at 13:51
8

$\exp$, $\sin$, $\cos$, $\log$... are solutions of very simple differential equations. This fact plus the existence of explicit rules for the derivatives of sum/difference/product/quotient/composition guarantee that no "new" functions will appear when deriving elementary functions.

Martín-Blas Pérez Pinilla
  • 41,546
  • 4
  • 46
  • 89
  • 1
    These functions are also solutions of very simple integral equations. There are also some rules for integration (not much, admittedly), but why don't they guarantee that no new function appears ? And why are the elementary functions solutions of simple differential equations ? ... –  Mar 20 '17 at 11:28
  • 2
    @YvesDaoust, (1) the condition is necessary but not sufficient. (2) no rule for integral of product to begin with. (3) $\exp$/$\log$ are homomorphisms ($\log$ was defined to be a homomorphism). This algebraic condition implies what will be its derivatives. A similar (geometric) reasoning can be done for $\sin$/$\cos$. And in the last pass to calculate explicitly the derivative of an elementary function you have to use that $\exp' = \exp$... – Martín-Blas Pérez Pinilla Mar 20 '17 at 11:41
  • Integration by part is a rule for the product. It expresses the integral of a product of functions in terms of another integral. The product rule for derivatives does nothing else. –  Mar 20 '17 at 11:47
  • But why does the homomorphism extends to the derivative and not the antiderivative ? –  Mar 20 '17 at 11:48
  • 1
    @YvesDaoust, the rule of product for $fg$ gives $(fg)'$ in function of $f'$, and $g'$, not another function. If you know $f'$ an $g'$, then you know $(fg)'$ always. – Martín-Blas Pérez Pinilla Mar 20 '17 at 11:50
  • $\int fg=f\int g-\int(f'\int g)$ not another function. So what ? –  Mar 20 '17 at 11:52
  • 1
    @YvesDaoust, $\int(f'\int g)$ requires a rule for the integral of a product. And for your question about homomorphism, I repeat (1). – Martín-Blas Pérez Pinilla Mar 20 '17 at 11:54
0

Elementary functions are linear combination or quotient of polynomial and exponential functions (yes, that include sin, cos, tan). No doubt polynomials has derivative to be polynomials. And exponential is the single most important function in the universe, and clearly it is always differentiable and produces only exponential functions. The whole process of taking derivative is a combination of above process, so they are certainly elementary.

Violapterin
  • 1,725
0
  • for the same reason that there are strict rules for grammar, but not for creating poems. Differentiation goes ‘downwards’ – in the direction of gravity, so to speak – whereas integration goes ‘upwards’ – against the pull of gravity, so to speak. The price of defeating gravity is ambiguity, which takes the form of a lack of strict rules. Notice that integration gives you a taste of ambiguity from the very start: changing the value of a function a single point does not change the value of the integral.
Mike Jones
  • 4,460